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Verbitsky }} \renewcommand{\@evenhead}{\@oddhead} \renewcommand{\@oddfoot}{\hfil\thepage\hfil} \renewcommand{\@evenfoot}{\@oddfoot}} \pagestyle{verbit} \setlength\paperheight {10in}% \setlength\paperwidth {13.5in} \setlength{\textwidth}{0.8\paperwidth} \setlength{\textheight}{0.8\paperheight} \setlength{\pdfpageheight}{\paperheight} \setlength{\pdfpagewidth}{\paperwidth} \addtolength{\topmargin}{-20mm} \addtolength{\leftmargin}{-25mm} \addtolength{\rightmargin}{-25mm} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % Lemma, sublemma, corollary, proposition, theorem, % % definition,example defined there: % %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \newcounter{section} \newcounter{Mycounter}[section] \newcounter{lemma}[section] \setcounter{lemma}{0} \renewcommand{\thelemma}{\noindent{Lemma \thesection.\arabic{lemma}}} \newcommand{\lemma}{% \setcounter{lemma}{\value{Mycounter}} \refstepcounter{lemma} \stepcounter{Mycounter} {\bf \green LEMMA:\ }} \newcounter{claim}[section] \setcounter{claim}{0} \renewcommand{\theclaim}{\noindent{Claim \thesection.\arabic{claim}}} \newcommand{\claim}{% \setcounter{claim}{\value{Mycounter}} \refstepcounter{claim} \stepcounter{Mycounter} {\bf \green CLAIM:\ }} \newcounter{corollary}[section] \setcounter{corollary}{0} \renewcommand{\thecorollary}{\noindent{Corollary \thesection.\arabic{corollary}}} \newcommand{\corollary}{% \setcounter{corollary}{\value{Mycounter}} \refstepcounter{corollary} \stepcounter{Mycounter} {\bf \green COROLLARY:\ }} \newcounter{theorem}[section] \setcounter{theorem}{0} \renewcommand{\thetheorem}{\noindent{Theorem \thesection.\arabic{theorem}}} \newcommand{\theorem}{% \setcounter{theorem}{\value{Mycounter}} \refstepcounter{theorem} \stepcounter{Mycounter} {\bf \green THEOREM:\ }} \newcounter{conjecture}[section] \setcounter{conjecture}{0} \renewcommand{\theconjecture}{\noindent{Conjecture \thesection.\arabic{conjecture}}} \newcommand{\conjecture}{% \setcounter{conjecture}{\value{Mycounter}} \refstepcounter{conjecture} \stepcounter{Mycounter} {\bf \green CONJECTURE:\ }} \newcounter{proposition}[section] \setcounter{proposition}{0} \renewcommand{\theproposition} {\noindent{Proposition \thesection.\arabic{proposition}}} \newcommand{\proposition}{% \setcounter{proposition}{\value{Mycounter}} \refstepcounter{proposition} \stepcounter{Mycounter} {\bf \green PROPOSITION:\ }} \newcounter{definition}[section] \setcounter{definition}{0} \renewcommand{\thedefinition} {\noindent{Definition~\thesection.\arabic{definition}}} \newcommand{\definition}{% \setcounter{definition}{\value{Mycounter}} \refstepcounter{definition} \stepcounter{Mycounter} {\bf \green DEFINITION:\ }} \newcounter{example}[section] \setcounter{example}{0} \renewcommand{\theexample}{\noindent{Example \thesection.\arabic{example}}} \newcommand{\example}{% \setcounter{example}{\value{Mycounter}} \refstepcounter{example} \stepcounter{Mycounter} {\bf \green EXAMPLE:\ }} \newcounter{remark}[section] \setcounter{remark}{0} \renewcommand{\theremark}{\noindent{Remark \thesection.\arabic{remark}}} \newcommand{\remark}{% \setcounter{remark}{\value{Mycounter}} \refstepcounter{remark} \stepcounter{Mycounter} {\bf \green REMARK:\ }} \newcounter{observation}[section] \setcounter{observation}{0} \renewcommand{\theobservation}{\noindent{Question \thesection.\arabic{observation}}} \newcommand{\observation}{% \setcounter{observation}{\value{Mycounter}} \refstepcounter{observation} \stepcounter{Mycounter} {\bf \green OBSERVATION:\ }} \newcounter{question}[section] \setcounter{question}{0} \renewcommand{\thequestion}{\noindent{Question \thesection.\arabic{question}}} \newcommand{\question}{% \setcounter{question}{\value{Mycounter}} \refstepcounter{question} \stepcounter{Mycounter} {\bf \green QUESTION:\ }} \newcommand{\exercise}{% {\bf \green EXERCISE:\ }} \newcommand{\proof}{% {\bf \green Proof:\ }} \newcommand{\pstep}{% {\bf \green Proof. Step 1:\ }} \begin{document} \setcounter{page}{1} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \begin{center} {\Large\bf Complex geometry\\[15mm] \small lecture 14: Dolbeault cohomology of a torus} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \\[14mm] {\small Misha Verbitsky } \\[20mm] {\tiny\bf HSE, room 306, 16:20, \\[2mm] November 7, 2020 } \end{center} \newpage {\bf \blue Weight decomposition for $T^n$-action (reminder)} \exercise Consider the $n$-dimensional torus $T^n$ as a Lie group, $T^n = U(1)^n$. {\bf \purple Prove that any finite-dimensional Hermitian representation of $T^n$ is a direct sum of 1-dimensional representations,} with action of $T^n$ given by $\rho(t_1, ..., t_n)(x)= \exp(2\pi\1\sum_{i=1}^np_it_i) x$, for some $p_1, ..., p_n\in \Z^n$, called {\bf \blue the weights} of the 1-dimensional representation. \definition Let $V$ be a Hermitian space (possibly infinitely-dimensional) equipped with an action of $T^n$, and $V_\alpha\subset V$ weight $\alpha$ representations, $\alpha\in \Z^n$. The direct sum $\bigoplus_{\alpha\in \Z^n} V_\alpha$ is called {\bf \blue the weight decomposition} for $V$ if it is dense in $V$. \theorem Let $W$ be a Hermitian vector space. Then {\bf \red the Fourier series provide the weight decomposition on $L^2(T^n, W)$.} \endproof \theorem Let $W$ be a Hermitian representation of $T^n$. {\bf \red Then $W$ admits a weight decomposition $V = \widehat{\bigoplus_{\alpha\in \Z^n} W_\alpha}$.} \proof We realize $W$ as a subrepresentation in $L^2(T^n, W)$, and use the Fourier series to obtain the weight decomposition of $L^2(T^n, W)$. \endproof \newpage {\bf \blue Weight decomposition for $T^n$-action on differential forms (reminder)} \remark Let $M$ be a manifold with the $T^n$-action, and \[ \Lambda^*(M)= \hat\bigoplus_{\alpha \in \Z^n} \Lambda^*(M)_{p_1, ..., p_k}\] be the weight decomposition on the differential forms. Then the {\bf \red de Rham differential preserves each term $\Lambda^*(M)_{p_1, ..., p_k}$.} Indeed, {\bf \purple $d$ commutes with the action of the Lie algebra of $T^n$, and $\Lambda^*(M)_{p_1, ..., p_k}$ are its eigenspaces.} \remark Let $\alpha= \sum \alpha_{p_1, ..., p_k}$ be the weight decomposition. {\bf \purple The forms $\alpha_{p_1, ..., p_k}$ are obtained by averaging \[ e^{2\pi\1\sum_{i=1}^np_it_i}\alpha = \Av_{T^n} e^{2\pi\1\sum_{i=1}^n-p_it_i}\alpha \] hence they are smooth.} \newpage {\bf \blue De Rham cohomology and $T^n$-action (reminder)} \theorem Let $M$ be a smooth manifold, and $T^n$ a torus acting on $M$ by diffeomorphisms. Denote by $\Lambda^*(M)^{T^n}$ the complex of $T^n$-invariant differential forms. {\bf \red Then the natural embedding $\Lambda^*(M)^{T^n}\hookrightarrow \Lambda^*(M)$ induces an isomorphism on de Rham cohomology.} \pstep Let $\alpha\in \Lambda^*(M)$ be a form and $\alpha= \sum \alpha_{p_1, ..., p_n}$ its weight decomposition, with $\alpha_{p_1, ..., p_n}\in \Lambda^*_{p_1, ..., p_n}(M)$ a form of weight $p_1, ... , p_n$. Since $T^n$-action commutes with de Rham differential, these forms are closed when $\alpha$ is closed. {\bf \green Step 2:} Let $r_1, ..., r_n$ be the standard generators of the Lie algebra of $T^n$ rescaled in such a way that $\Lie_{r_k}(\exp(2\pi\1\sum_{i=1}^np_it_i))=\1 p_k$, and $i_{r_k}:\; \Lambda^i(M)\arrow \Lambda^{i-1}(M)$ the convolution operator. Since $\Lie_{r_k}= \{d, i_{r_k}\}$, we have $p_k\alpha_{p_1, ..., p_n}= d (i_{r_k}\alpha_{p_1, ..., p_n})$ whenever $\alpha_{p_1, ..., p_n}$ is closed. Therefore, {\bf \purple all terms in the weight decomposition $\alpha= \sum \alpha_{p_1, ..., p_n}$ are exact except $\alpha_{0,0, ...,0}$.} {\bf \green Step 3:} In the direct sum decomposition of the de Rham complex \[ \Lambda^*(M)= \Lambda^*(M)^{T^n}\oplus \hat\bigoplus_{p_1, ..., p_k\neq (0,0,..., 0)}\Lambda^*_{p_1, ..., p_k}(M) \] the second component has trivial cohomology, because $\Lie_{r_k}$ is invertible on $\bigoplus_{p_k\neq 0}\Lambda^*_{p_1, ..., p_n}(M)$ {\bf \purple (deduce it from $p_k\alpha_{p_1, ..., p_k}= d (i_{r_k}\alpha_{p_1, ..., p_k})$),} and\\ $\Lie_{r_k}(\text{closed form})$ is exact. \endproof \newpage {\bf \blue Constant forms on a torus} \remark In the proof above, the serie $\sum_{p_k\neq 0} \frac 1 {p_k}i_{r_k}\alpha_{p_1, ..., p_k}$ converges, because $\sum_{p_k\neq 0}\alpha_{p_1, ..., p_k}$ converges, and satisfies $d\left( \sum_{p_k\neq 0} \frac 1 {p_k}i_{r_k}\alpha_{p_1, ..., p_k}\right) =\sum_{p_k\neq 0}\alpha_{p_1, ..., p_k}$ as shown. \definition Let $T^n=(S^1)^n$ be a compact torus equipped with a action on itself by shifts, and $\Lambda^*_\const(M)$. the space of $T^n$-invariant forms on $T^n$. These forms are called {\bf \blue constant differential forms}. Clearly, {\bf \purple constant forms have constant coefficients in the usual (flat) coordinates on the torus.} \theorem The natural embedding $\Lambda^*_\const(T^n)\hookrightarrow \Lambda^*(T^n)$ {\bf \red induces an isomorphism $\Lambda^*_\const(T^n)=H^*(T^n)$.} \proof The embedding $\Lambda^*_\const(T^n)= \Lambda^*(T^n)^{T_n} \hookrightarrow \Lambda^*(T^n)$ induces an isomorphism on cohomology, however, all constant forms are closed, hence $H^*(\Lambda^*_\const(T^n), d)=\Lambda^*_\const(T^n)$. \endproof \newpage {\bf \blue Holomorphic vector fields} \definition Let $(M,I)$ be a complex manifold, and $X\in TM$ a real vector field. It is called {\bf \blue holomorphic} if $\Lie_X(I)=0$, that is, if the corresponding flow of diffeomorphisms is holomorphic. \claim Let $(M,I)$ be a complex manifold, and $X\in TM$ a holomorphic vector field. {\bf \red Then $X^c:= I(X)$ is also holomorphic, and commutes with $X$.} \pstep Assume that $X$ is non-zero at a given point $m\in M$ Solving the appropriate differential equation in holomorphic coordinates, we obtain a coordinate system $z_1, ..., z_n$ in a neighbourhood of $m$ such that $\Lie_X z_i= 0$ for $i>1$ and $\Lie_{z_1}=1$. Let $x_i, y_i$ be the corresponding real coordinate system, wih $x_i =\Re z_i$ and $y_i =\Im z_i$. Then $X= \frac d{dx_1}$ and $X^c= \frac d{dy_1}$. {\bf \green Step 2:} The conditions $\Lie_{X^c}(I)=0$ and $[X^c,X]=0$ hold on a closed subset of $M$, that is, they are true on the closure $C$ of the set of points where $X\neq 0$. Outside of $C$, the vector field $X$ is identically zero, hence these conditions are also hold. \endproof \newpage {\bf \blue Cartan's formula for Dolbeault differential} \lemma Let $X$ be a holomorphic vector field, and $X^c=I(X)$. {\bf \red Then $\{d^c, i_X\}= - \Lie_{X^c}$.} \proof Using $\{IdI^{-1}, i_X\}= I\{d, I^{-1}i_XI\}I^{-1}$, we obtain $\{d^c, i_X\}= - I \{d, i_{X^c}\}I^{-1}= I\Lie_{X^c}I^{-1}$. However, $X^c$ is holomorphic, hence $I\Lie_{X^c}I^{-1}= \Lie_{X^c}$. \endproof \proposition Let $X$ be a holomorphic vector field, and $X^c=I(X)$. {\bf \red Then $\{\bar\6, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})$.} \proof $\bar\6= \frac 1 2 (d +\1 d^c)$, hence \[ \{\bar\6, i_X\}= \frac 1 2 \Lie_X + \1 \{d^c, i_X\}= \frac 1 2 (\Lie_X - \1 \Lie_{X^c}). \] \endproof \remark Let $M$ be a complex manifold equipped with a holomorphic action of the torus $T^n$. Then the action of $T^n$ commutes with $d$ and $d^c$. Therefore, the operators $d, d^c$ preserve the eigenspaces of the corresponding Lie algebra. These eigenspaces are components of the weight decomposition. This implies that {\bf \purple the Dolbeault differential $\bar\6$ preserves the weight decomposition.} \newpage {\bf \blue Dolbeault cohomology of an elliptic curve} \definition {\bf \blue An elliptic curve} is a 1-dimensional compact complex manifold $X:=\C/\Z^2$. \remark The additive group $\C$ acts on itself by parallel transforms, hence {\bf \red the 2-dimensional torus $T^2$ acts on an elliptic curve by holomorphic diffeomorphisms.} \definition The $T^n$-invariant forms on $T^n$ are called {\bf \blue constant}. \definition {\bf \blue Dolbeault cohomology} of a complex manifold is $\frac{\ker \bar\6}{\im\bar\6}$. \corollary {\bf \red Dolbeault cohomology of an elliptic curve $X$ are represented by the constant forms on $X$.} {\bf \green Proof using the Hodge theory:} Choose a $T^2$-invariant K\"ahler form on $X$. We have already obtained an isomorphism between de Rham cohomology and the constant forms. Since the constant forms are harmonic, there are no other harmonic forms. Now, $\Delta_{\bar\6}=\frac 1 2 \Delta_d$, hence\\ \phantom{x} \ \ \ \ \ \ constant forms = $\bar\6$-harmonic forms = Dolbeault cohomology. \endproof In the next slide, {\bf \green we give a proof which is independent from the Hodge theory.} \newpage {\bf \blue Dolbeault cohomology of an elliptic curve (2)} \proposition Let $X$ be an elliptic curve, and $\Lambda^*(X)= \bigoplus_{\alpha \in \Z^2} \Lambda^*(X)_{p_1,p_2}$ its weight decomposition under the $T^2$-action. Consider the space $T^2$-invariant forms $\Lambda^*(X)^{T^2}=\Lambda^*(X)_{0,0}$. {\bf \red Then the natural embedding $\Lambda^*(X)^{T^2} \hookrightarrow \Lambda^*(X)$ induces an isomorphism of Dolbeault cohomology.} \proof Let $\alpha\in \Lambda^*(X)_{p_1,p_2}$ be a $\bar\6$-closed form, with $(p, q)\neq (0,0)$. Suppose, for example, that $p\neq 0$, and $X$ is the generator of the corresponding component of the Lie algebra such that $\Lie_X\alpha=p\1\alpha$. Since $X^c$ belongs to the same Lie algebra, we have $\Lie_{X^c}(\alpha)= v\alpha$, where $v\in \1\R$. Then \[ \frac{\1p+ v}{2} \alpha= \frac 1 2 (\Lie_X - \1 \Lie_{X^c})\alpha=\{ \bar\6, i_X\}\alpha=\bar\6 i_X\alpha, \ \ \ (***) \] hence $\alpha$ is $\bar\6$-exact. This implies that {\bf \purple $\bar\6$ has no cohomology on \[ \bigoplus_{p_1, p_2\neq (0,0)}\Lambda^*(X)_{p_1,p_2}.\]} \endproof \newpage {\bf \blue $\bar\6$-exact top forms on an elliptic curve} \claim Let $\eta\in \Lambda^n(T^n)$ be a top form on a torus, and $\nu =\sum_{\alpha\in \Z^n} \nu_\alpha$ its weight decomposition. {\bf \red Then $\int_{T^n} \nu = \int_{T^n} \nu_0$, where $\nu_0$ denotes the $T^n$-invariant component.} Moreover, {\bf \red whenever $\int_{T^n} \nu=0$, the component $\nu_0$ also vanishes.} \proof Let $\nu$ be a top form on a compact manifold, equipped with an action of $S^1$, and $\nu =\sum \nu_i$ its weight decomposition. {\bf \purple Then $\int_M \nu_i=0$ for all $i\neq 0$.} Indeed, the $S^1$-action multiplies $\nu_i$ by a non-zero number, but the integral is invariant under the action of diffeomorphisms. \endproof \proposition Let $\eta\in \Lambda^2(X)$ be a form on an elliptic curve such that $\int_X\eta=0$. {\bf \red Then $\eta$ is $\bar\6$-exact.} \proof Consider the weight decomposition $\eta =\sum_{\alpha\in \Z^2} \eta_\alpha$. Since $\int_M \eta=0$, the (0,0)-component vanishes, and by (***) the form $\eta$ is $\bar\6$-exact. \endproof \newpage {\bf \blue Dolbeault cohomology of a disk} \corollary Let $K\subset \C$ be a compact subset, $K^0$ its interior, and $\eta\in \Lambda^2(K^0)$ a top form smoothly extending to a neighbourhood of $K$. {\bf \red Then $\eta$ is $\bar\6$-exact.} \proof Choosing an appropriate lattice $\Z^2\subset \C$, we may assume that $K$ is a subset of an elliptic curve $X$. Since $\eta$ extends to a neighbourhood of $K$, we can use partition of unity to extend it to a smooth form $\tilde \eta$ on $X$. Applying the weight decomposition $\tilde \eta =\sum_{\alpha\in \Z^2} \eta_\alpha$, we obtain that the form $\eta- \eta_{0,0}$ is $\bar\6$-exact. However, the constant part $\eta_{0,0}=\const\cdot dz \wedge d\bar z= \const \cdot\bar\6(\bar z dz)$ is also $\bar\6$-exact. \endproof \corollary Let $K\subset \C$ be a compact subset, $K^0$ its interior, and $\mu\in \Lambda^2(K^0)$ a (0,1)-form smoothly extending to a neighbourhood of $K$. {\bf \red Then $\mu$ is $\bar\6$-exact.} \proof By the previous corollary, $\mu\wedge dz$ is $\bar\6$-exact: there exists a (1,0)-form $\phi$ such that $\bar\6\phi =\mu\wedge dz$. However, for any (1,0)-form $\phi$ there exists a function $\psi$ such that $\psi dz= \phi$, which gives $\bar\6\psi =\mu$ because the map $\Lambda^{0,1}(X)\stackrel {\wedge dz}\arrow \Lambda^{1,1}(X)$ is an isomorphism which commutes with $\bar\6$. \endproof \newpage {\bf \blue Poincar\'e-Dolbeault-Grothendieck lemma} \definition {\bf \blue Polydisc} $D^n$ is a product of $n$ discs $D\subset \C$. \theorem {\bf \blue (Poincar\'e-Dolbeault-Grothendieck lemma)}\\ Let $\eta \in \Lambda^{p,q}(D^n)$, $q>0$, be a $\bar\6$-closed form on a polydisc, smoothly extended to a neighbourhood of its closure $\overline{D^n}\subset \C^n$. {\bf \red Then $\eta$ is $\bar\6$-exact.} {\bf \green We proved it for $n=1$.} Nextr lecture we prove it for all $n$. \end{document}